Show f(x)=x^3 is 1-1 for U(16)

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In summary, the conversation discusses proving that the function f(x)=x^3 is an automorphism of U(16) with operation (multiplication) mod 16. The main difficulty is showing that f is one-to-one. It is suggested to calculate f(x) for all possible x values to prove this, although it may be tedious. Another approach is to observe that for f(x) and f(y) to be distinct elements of U(16), their difference must be divisible by 16, but this is not possible if x and y are distinct. The conversation also addresses the question of whether f(a) and f(b) being equal mod 16 implies that a and b are equal.
  • #1
1800bigk
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I need to show that f(x)=x^3 is an automorphism of U(16) ie.
{1,3,5,7,9,11,13,15} with operation (multiplication)mod 16. I am having trouble showing that f is 1 to 1. I know it is 1 to 1 because I took each element calculated it to make sure, but how do I show that it is 1 to 1. I would usually assume f(a)=f(b) then show a = b but I am stuck there. Once I show its 1 to 1 I am pretty much done because its guaranteed to be onto since U(16) is finite.
 
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  • #2
Showing that f(x)= f(y) only if x= y by calculating f(x) for every possible x is tedious but completely valid.
 
  • #3
Well if you calculated them all, and they're all different, that constitutes a proof, albeit a rather unelegant one. Suppose x and y are distinct. To show f(x) and f(y) are distinct, observe:

f(x) - f(y) = x³ - y³ = (x-y)(x² + xy + y²) = (A)(B)

i.e. A = x-y, B = x² + xy + y²

Now for f(x) and f(y) to be distinct elements of U(16), AB will have to be divisible by 16. But note that B is an odd number, since it is the sum of 3 numbers, each of which is odd because each of them is a product of two odd numbers. So A would itself would have to be divisible by 16 (possibly being 0). But A is not zero because we're trying to prove that when x and y are distinct, then f maps them to different elements. And A is certainly no other multiple of 16, because two distinct elements of that set can't possibly differ by 16.
 
  • #4
thanks, if a,b in U(16) does (a^3)mod16 = (b^3)mod16 imply (a^3)=(b^3)?
 

FAQ: Show f(x)=x^3 is 1-1 for U(16)

What does it mean for a function to be one-to-one?

A one-to-one function, also known as an injective function, is a function in which each element of the domain maps to a unique element in the range. In other words, no two different inputs can result in the same output.

How can you determine if a function is one-to-one?

There are a few ways to determine if a function is one-to-one. One way is to graph the function and see if it passes the horizontal line test, meaning that no horizontal line intersects the graph more than once. Another way is to use the vertical line test, where if every vertical line intersects the graph at most once, the function is one-to-one. Lastly, for a function f(x) to be one-to-one, it must satisfy the following property: if f(a) = f(b), then a = b.

What is the domain and range of the function f(x)=x^3?

The domain of f(x)=x^3 is all real numbers, as any value can be cubed. The range of f(x)=x^3 is also all real numbers, as any number can be the result of cubing another number.

Can the function f(x)=x^3 be one-to-one for all values of x?

No, the function f(x)=x^3 is not one-to-one for all values of x. In fact, for any function that is a polynomial of degree greater than 1, there will be values of x that result in the same output. In the case of f(x)=x^3, x=0 and x=-1 both result in an output of 0, making it not one-to-one for those values.

How can you prove that the function f(x)=x^3 is one-to-one for U(16)?

To prove that the function f(x)=x^3 is one-to-one for U(16), we need to show that for any two different inputs a and b, the outputs f(a) and f(b) will also be different. For U(16), the domain is limited to the numbers 1, 2, 3, ..., 15, 16, and since these are all different numbers, their cubes will also be different. Therefore, f(x)=x^3 is one-to-one for U(16).

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